CAT4MBA

Q1) What are the last two digits of 122 x 123 x 125 x 127 x 129? Ans = 50

Q2)What are the last two digits of 15 x 37 x 63 x 51 x 97 x 17? Ans = 35

Q3) What will be the LAST THREE DIGITS in above qns?

 plz help....

Q1) What are the last two digits of 122 x 123 x 125 x 127 x 129?

solution: ny multiple of 25 should end in 25or50or75or00

furthr an odd  multiple of 25 n 2 ends in 50

n even multiple of 25*2 ends in 00

hence 50

Hi,

    Here is my work-put....Plz post the options so tht we can determine the ans from t.

6^anyting ends with 6

so,

    the equ boils down to 8^7^6

7^6 ends with 9

8^9 ends with 8.

All the numbers/8 must be divisible by 4.so the last 2 digits must be divisible by 4.

so it shud be any1 of the below:

.......08

.........28

...........48

............68

.............88

So plz post the options..

-Deepak.

Hey, Your solution is wrong.  The problem simplifies as 8^7^(a no. ending in 6). You've wrongly assumed that it is 7^6 and thus is ending in 9. It might be 7^(.....16) or 7^(......26) etc . Thus it might not end in 9. Thus ur solution is wrong...

I am not sure if the following is 100% correct and would love to see any flaw in the method.

Any how I would never attempt this sort of questions in real CAT.

3² = 9   

4⁹ = Something

5^Something = x x x x. . . .25

Next target 6 ^{x x x x. . . .25}  

6¹ = 6

6² = 36

6³ = X16

6⁴ =XX96

6⁵ = XXX76

6⁶ = XXXXX56

6⁷ = XXXXXX36

Last two digits periodicity = 5

Now x x x . . . . 25 is divisible by 5.

Thus 6 ^{x x x x. . . .25}   = XXXXX. . . . .76

Next target 7^{XXXXX. . . . .76}

7¹ = 7

7² = 49

7³ =  X43

7⁴ = XX01

7⁵ = XX07

Last two digits periodicity for 7 = 4

and 76 is divisible by 4
7^{XXXXX. . . . .76  =}  XXXXXXXX. . . . .01

Final target 8^{XXXXX. . . . .01}

Last two digits periodicity for 8 = 20

64        44          24          04         84                64

12        52          92          32          72

96        16          36          56           76

68        28          88          48             8

So 8^{XXXXX. . . . .01} ends with 08 and thats theanswer

Wonderfully done, man !!!....I was thinking along the same lines but did not carry on till the end....only one small thing I would like to point out that while finding the periodicity of the last 2 digits for 8, you could have done with only 3 columns of calculation,

64        44          24  ....    

12        52          92  ....

96        16          36  ....      

68        28          88  ....

     
since in each row the tens digit increases or decreases by the same amount in consecutive columns and comes back to the original value after a cycle of 5 columns (in all the rows). So we can conclude that the periodicity is 20.

But I totally agree with you that we shudn't attempt such questions in cat...

good funda

is the answer of the 4th question is     01

last two digits of 21^100 is 01 please correct me if m wrong

guys help me solving this question   

what is the remainder when 32^33^34 is divided by 7

a-4   b-2   c-3   d-1

reminder of 32³³³⁴divided by 7

= reminder of 4³³³⁴divided by 7

= reminder of 4^3K divided by 7

4 divided by 7 has a cylicity of 3 and the reminders are 4, 2, 1

So the answer is 1